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4 votes
For normally distributed data, what proportion of observations have z-scores satisfying the following conditions:

0.62 < z < 2.92

User Breen Ho
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1 Answer

4 votes
From standard tables for normal distribution,
P(z<=0.62) = 0.7324
P(z<=2.92) = 0.9983

Therefore
P(0.62 < z <2.92) = 0.9983 - 0.7324 = 0.2659

The required proportion is 0.2659 or 26.6%

Answer: 0.266 or 26.6%
User Mike Kaply
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8.3k points